On the straight line passing through points A (11; 15) and B (2; 3) find the ordinate of point C, the abscissa of which is 5.

Let us write the canonical equation of the straight line that passes through the given two points

A (11; 15) and B (2; 3):

(x – 11) / (2 – 11) = (y – 15) / (3 – 15);

(x – 11) / (-9) = (y – 15) / (-12);

(x – 11) / 3 = (y – 15) / 4;

4 * x – 44 = 3 * y – 45;

4 * x – 44 – 3 * y + 45 = 0;

4 * x – 3 * y +1 = 0. …

Point C (5; y) lies on this straight line. Substitute the straight line x into the equation and find y:

4 * 5 – 3 * y + 1 = 0;

3 * y = 21;

y = 7.

Answer: the ordinate of point C is 7,